Porisma

Worked solution

Distinguish six separate rods from three times one rod

Type A rod lengths are normally distributed with mean 5 cm and standard deviation 0.5 cm; type B lengths have mean 10 cm and standard deviation 1 cm. Find the probability that

(a) the total length of 3 type A rods and 2 type B rods lies between 33 and 36 cm,

(b) the total length of 6 type A rods exceeds three times the length of one randomly chosen type B rod.

CJC · Discussion · Question 8

Answer

(a)0.613 (3 s.f.); (b)1/2.

Define independent rod observations

Let A∼N(5,0.52)A\sim N(5,0.5^2) and B∼N(10,12)B\sim N(10,1^2) in cm. The source adds variances on the assumption that distinct randomly selected rods have independent lengths. This assumption is necessary for its sum models.

How did this part go?

(a) Model the five-rod total

A1+A2+A3+B1+B2∼N(3(5)+2(10),3(0.52)+2(12))=N(35,2.75)A_1+A_2+A_3+B_1+B_2\sim N(3(5)+2(10),3(0.5^2)+2(1^2))=N(35,2.75). Thus P(33<A1+A2+A3+B1+B2<36)=0.613P(33<A_1+A_2+A_3+B_1+B_2<36)=0.613 to 3 significant figures.

How did this part go?

(b) Combine independent rods with one scaled rod

A1+⋯+A6−3B∼N(6(5)−3(10),6(0.52)+32(12))=N(0,10.5)A_1+\cdots+A_6-3B\sim N(6(5)-3(10),6(0.5^2)+3^2(1^2))=N(0,10.5). The six A rods contribute six variances; the single B rod is multiplied by 3 and contributes nine times its variance.

How did this part go?

(b) Use symmetry about zero

The required probability is P(A1+⋯+A6−3B>0)=1/2P(A_1+\cdots+A_6-3B>0)=1/2, because this continuous normal difference has mean 0.

How did this part go?

Common mistakes

Three times one B length is not the sum of three independent B lengths.

Solution notes

Both asks, source initial models, all independent sum/scaling mean/variance terms and symmetry conclusion retained. Independent selected-rod assumption stated explicitly; no marks printed.

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